Conservation of momentum and net force

[silversurf]

does anyone know why momentum is not conserved in this problem? I thought momentum was always conserved when no net force acted on the system? Is the ground considered the net force acting on the system? Even though it changes direction I was thinking momentum would still be conserved except that the ball would move in the opposite direction.

This is the question.

Which of the following quantities is (are) conserved when a falling object strikes the ground?

I.


Momentum of the object

II.


Kinetic energy of the object

III.


Total energy





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Correct Answer:



III only


Your answer: You were incorrect



I and III only



III only

Explanation



I.


Momentum of the object

II.


Kinetic energy of the object

III.


Total energy

Total energy (III.) is conserved when a falling object strikes the ground.



Since the object’s velocity drops to zero after the impact, its momentum certainly changes (eliminating I only and I and III only) and its kinetic energy drops to zero (eliminating II and III only). The total energy of a system (which includes heat) remains constant.

 

[Simon Bridge]

Welcome to PF;
You appear to have answered your own question in the final statement:

Since the object’s velocity drops to zero after the impact, its momentum certainly changes (eliminating I only and I and III only) and its kinetic energy drops to zero (eliminating II and III only). The total energy of a system (which includes heat) remains constant.

... and yet you seem to be confused still?

The way to understand conservation problems is to clearly write down the situation "before" and the situation "after" before comparing them.

What is the total momentum before the collision?
What is the total momentum after the collision?
(Say the object has mass m and initial velocity v, and the ground has mass M - what are it's initial and final velocities?)

 

[WannabeNewton]

The momentum during the entire trajectory is not conserved. This is because the object is free falling before it actually hits the ground i.e. it is under a force $F = \frac{dp}{dt} = -mg\neq 0$ during that part of its trajectory. However the momentum is conserved during the impact itself, which only lasts for an infinitesimal amount of time. Let the ground have some finite mass $M$ then we have $\frac{mv_{i}}{(M + m)} = v_{f}$ where $v_i$ is the velocity of the falling object immediately before striking the ground and $v_f$ is the velocity of the combined ground + object system immediately after the collision. What we then do is take the limit as $M$ becomes extremely large so that $v_{f}$ is vanishingly small.

The reason we neglect the effect of gravity during the infinitesimal time over which the collision occurs is because the change in momentum due to gravity over that time interval will be $\Delta p = -mg \Delta t\approx 0$ since $\Delta t$ is itself vanishingly small and $-mg$ is constant.

EDIT: I see Simon beat me to it. This is the second time today dagnabbit!

 

[Simon Bridge]

I think the problem implies that the reference frame is attached to the ground ... don't worry, I'm off to work soon ;)

 

[silversurf]

Hey guys thanks for your feedback. I think I got it. I was confused because I was looking at the falling object only. So I was thinking of a ball hitting the ground and then bouncing back up with the same momentum but in the opposite direction and wasn't taking into account the mass of the ground as being part of the system. So is this reasoning correct? The the momentum changes because the ground (external net force) acts on our system and changes it's momentum to pretty much zero because the huge increase in mass of our system completely stops our object from bouncing back up as it drops the final velocity to zero?

 

[rcgldr]

does anyone know why momentum is not conserved in this problem?

Momentum is conservered if you include the effects on the earth. Since the Earth is massive, the change in Earth's velocity due to the object striking the ground will be very tiny, but momentum of the Earth and object will be conserved.

 

[silversurf]

Momentum is conservered if you include the effects on the earth. Since the Earth is massive, the change in Earth's velocity due to the object striking the ground will be very tiny, but momentum of the Earth and object will be conserved.

From the way the question was worded, why would we assume that we don't include the effects of gravity?

 

[Simon Bridge]

I was looking at the falling object only. So I was thinking of a ball hitting the ground and then bouncing back up with the same momentum but in the opposite direction

The question says nothing about the object bouncing back - just says it hits the ground. What if the ground is soft and the ball sticks there? Does that make a difference?

There are a great many different ways the ground and the ball can interact besides the ones you have considered. When you answer a question like that you need to make judgement calls about what it means.

 

[cloa513]

I am not sure if anyone mentioned but two of the answers only mention the object- its properties alone are not conserved-　the system properties- object plus Earth are reasonably conserved (everything is affected by forces e.g. gravity).

 

[Viracocha]

Which of the following quantities is (are) conserved when a falling object strikes the ground?

I.


Momentum of the object

II.


Kinetic energy of the object

III.


Total energy

It's really more about wording. It says "the object", of course the momentum and KE of the object would change after a collision, although the total energy remains constant

 

[Simon Bridge]

Oh yes - that's how I was reading it.
The trouble with this is that many courses develop their own shorthand in context, so we cannot be sure.